Good morning,

I want to find a solution to the following equation:

$$\left|u\cdot\sin\left(2\pi\cdot f\cdot t\right)\right|=u\cdot\exp\left(-\frac{t}{a}\right)\space\Longleftrightarrow\space t=\dots\tag1$$

The sine-function is a periodic function, so I want only one solution and that is the solution that is between the follwing boundaries:

$$\frac{1}{2\cdot f}<t<\frac{3}{4\cdot f}\tag2$$

Now, alle the variables are real numbers and bigger than zero.

Question: How can Mathematica 10.0 solve this question? Or approximate the solution?


With new variables/parameters ft=f t, fa=f a, assuming u>0, you have to solve the equation -Sin[2 Pi ft] == Exp[ -ft/fa] ,1/2 < ft < 3/4.

sol[fa_?NumericQ] :=NMinimize[{1, {-Sin[2 Pi ft] == Exp[ -ft/fa], 1/2 < ft <3/4}},ft  ][[2]]     

sol[.5] (*{ft -> 0.553605} *)
Plot[ ft /. sol[fa] , {fa, 0, 10}, PlotRange -> {0, Automatic},AxesLabel -> {f a, f t}]

enter image description here

  • $\begingroup$ Well, I need to find a general code where I know $\text{f}$ and $\text{a}$ and I need to solve for $t$. I don't need to know how to plot $\endgroup$ Feb 9 '18 at 8:09
  • $\begingroup$ sol[]is the numerical function you asked for! The adaption on your parameters f, a, t is easily done ... $\endgroup$ Feb 9 '18 at 8:12
  • $\begingroup$ sol[] is not a function in Mathematica 10.0. And why do you use $-\sin$? And $\text{f}$ and $t$ are two different things so writing $ft$ is a bit confusing. The only unknown in the equation is $t$ $\endgroup$ Feb 9 '18 at 8:14
  • 1
    $\begingroup$ sol[] is a user defined function I gave you in my code! In the given range of f *t you can substitude Abs[Sin[...]] by -Sin[]. Resubstituting ft by f*t, fa by f* a seems to be easily solvable... $\endgroup$ Feb 9 '18 at 8:21

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